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13. A catalyst lowers Ea for a reaction from 85.0 to 70.5 kJ molG1 at 27EC....

13. A catalyst lowers Ea for a reaction from 85.0 to 70.5 kJ molG1 at 27EC.
If the rate of the uncatalyzed reaction is 2.00 x 10 —3 mol L—1s —1, what
is the rate of the catalyzed reaction ?

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Answer #1

ln K1 = ln A - Ea1/RT

ln K2 = ln A - Ea2/RT

Subtract 1st one from 2nd:

ln K2 - ln K1 = Ea1/RT - Ea2/RT

ln K2/K1 = (Ea1-Ea2)/RT

Here:

Ea1-Ea2 = 85.0 - 70.5 = 14.5 KJ/mol = 1450.0 J/mol

T= 27.0 oC

= (27.0+273) K

= 300 K

use:

ln K2/K1 = (Ea1-Ea2)/RT

ln K2/K1 = (14500.0)/(8.314*300)

ln K2/K1 = 5.813

K2/K1 = e^(5.813)

K2/K1 = 3.348*10^2

K2 / (2.00*10^-3) = 3.348*10^2

K2 = 0.670 mol.L-1.s-1

Answer: 0.670 mol.L-1.s-1

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